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      <h1 id="Leecode-面试题40-最小的k个数"><a href="#Leecode-面试题40-最小的k个数" class="headerlink" title="Leecode-面试题40 最小的k个数"></a>Leecode-面试题40 <a href="https://leetcode-cn.com/problems/zui-xiao-de-kge-shu-lcof/" target="_blank" rel="noopener">最小的k个数</a></h1><h2 id="思路：快排-堆"><a href="#思路：快排-堆" class="headerlink" title="思路：快排/堆"></a>思路：快排/堆</h2><p><strong>题目描述</strong></p>
<p>输入整数数组 <code>arr</code> ，找出其中最小的 <code>k</code> 个数。</p>
<p><strong>示例 1：</strong></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入：arr &#x3D; [3,2,1], k &#x3D; 2</span><br><span class="line">输出：[1,2] 或者 [2,1]</span><br></pre></td></tr></table></figure>

<p><strong>示例 2：</strong></p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">输入：arr &#x3D; [0,1,2,1], k &#x3D; 1</span><br><span class="line">输出：[0]</span><br></pre></td></tr></table></figure>

<a id="more"></a>

<p><strong>Solution：</strong></p>
<ul>
<li><p>TopK问题，不管是求前K大/前K小/第K大/第K小等，都有3中不错的方法</p>
<p>​    <strong>1. O(N)：用快排变形最最最高效解决TopK问题</strong> </p>
<p>​    <strong>2.O (Nlogk):  大根堆（前K小）/小根堆（前K大）</strong></p>
<p>​    <strong>3. O(NlogK)：二叉搜索树</strong> </p>
</li>
</ul>
<h2 id="Java"><a href="#Java" class="headerlink" title="Java"></a>Java</h2><p><strong>Solution 1 快排:</strong></p>
<p>​    注意找前K大/前K小/第K大/第K小，是不需要对整个数组进行O(NlogN)的排序的！因为可以通过快排切分直接O(N)找到第K大的数，如果只会先排序再找的话，那啥…基本上就交代了叭( ͡° ͜ʖ ͡°)…）。</p>
<p>​    因此本题先通过快排切分排好第K小的数，根据快排切分的性质，它左边的k-1个数都小于等于</p>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span>&#123;</span><br><span class="line">    <span class="keyword">public</span> <span class="keyword">int</span>[] getLeastNumbers(<span class="keyword">int</span>[] arr, <span class="keyword">int</span> k) &#123;</span><br><span class="line">       <span class="keyword">if</span> (k == <span class="number">0</span> || arr.length == <span class="number">0</span>) &#123;</span><br><span class="line">            <span class="keyword">return</span> <span class="keyword">new</span> <span class="keyword">int</span>[<span class="number">0</span>];</span><br><span class="line">       &#125;</span><br><span class="line">       <span class="comment">//注意最后一个参数传入我们要找的下标（第k小的数下标是k-1）</span></span><br><span class="line">       <span class="keyword">return</span> quickSearch(arr,<span class="number">0</span>,arr.length - <span class="number">1</span>, k - <span class="number">1</span>);</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    <span class="keyword">private</span> <span class="keyword">int</span>[] quickSearch(<span class="keyword">int</span>[] nums,<span class="keyword">int</span> lo,<span class="keyword">int</span> hi,<span class="keyword">int</span> k)&#123;</span><br><span class="line">        <span class="comment">//每次快排切分一次，找到排序后小标为j的元素，如果j恰好等于k就返回j以及j左边所有的数</span></span><br><span class="line">        <span class="keyword">int</span> j = partition(nums,lo,hi);</span><br><span class="line">        <span class="keyword">if</span>(j == k)&#123;</span><br><span class="line">            <span class="keyword">return</span> Arrays.copyOf(nums,j + <span class="number">1</span>);</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> j &gt; k? quickSearch(nums,lo,j - <span class="number">1</span>,k) : quickSearch(nums,j + <span class="number">1</span>,hi,k);</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    <span class="comment">//快排切分，返回下标j，使得比num[j]小的数都在j的左边，比num[j]大的数都在j的右边</span></span><br><span class="line">    <span class="function"><span class="keyword">private</span> <span class="keyword">int</span> <span class="title">partition</span><span class="params">(<span class="keyword">int</span>[] nums, <span class="keyword">int</span> lo, <span class="keyword">int</span> hi)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">int</span> v = nums[lo];</span><br><span class="line">        <span class="keyword">int</span> i = lo, j = hi + <span class="number">1</span>;</span><br><span class="line">        <span class="keyword">while</span> (<span class="keyword">true</span>) &#123;</span><br><span class="line">            <span class="keyword">while</span> (++i &lt;= hi &amp;&amp; nums[i] &lt; v);</span><br><span class="line">            <span class="keyword">while</span> (--j &gt;= lo &amp;&amp; nums[j] &gt; v);</span><br><span class="line">            <span class="keyword">if</span> (i &gt;= j) &#123;</span><br><span class="line">                <span class="keyword">break</span>;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="keyword">int</span> t = nums[j];</span><br><span class="line">            nums[j] = nums[i];</span><br><span class="line">            nums[i] = t;</span><br><span class="line">        &#125;</span><br><span class="line">        nums[lo] = nums[j];</span><br><span class="line">        nums[j] = v;</span><br><span class="line">        <span class="keyword">return</span> j;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

<p><strong>快排变形时间复杂度分析：</strong></p>
<p>​    因为我们是要找下标为k的元素，第一次切分的时候需要遍历整个数组(0 ~ n)找到了下标是j的元素，假如k比j小的话，那么我们下次切分只要遍历数组(0~k-1)的元素就行啦，反之如果k比j大的话，那下次切分只要遍历数组(k+1～n)的元素就行啦，总之平均情况下，可以看作每次调用partition遍历的元素数目都是上一次遍历的1/2，因此时间复杂度是N + N/2 + N/4 + … + N/N = 2N, 因此时间复杂度是<strong>O(N)</strong>。</p>
<p><strong>Solution 2 堆:</strong></p>
<p>用堆虽然时间复杂度比快排变形慢了点，但是因为Java中提供了现成的PriorityQueue（默认小根堆），所以不需要自己写大段的模版代码，因此实现起来最简单，没几行代码，写起来很快～～面试的时候可以先快点写出这个方案🤫</p>
<p>   注意本题是求<strong>前K小</strong>，因此用一个容量为K的<strong>大根堆（</strong>每次poll出最大的数，那堆中保留的就是前K小啦）。注意不是小根堆嗷！小根堆的话需要把全部的元素都入堆，那是O(NlogN)😂，就不是O(NlogK)啦～～</p>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">// 保持堆的大小为K，然后遍历数组中的数字，遍历的时候做如下判断：</span></span><br><span class="line"><span class="comment">// 1. 若目前堆的大小小于K，将当前数字放入堆中。</span></span><br><span class="line"><span class="comment">// 2. 否则判断当前数字与大根堆堆顶元素的大小关系，如果当前数字比大根堆堆顶还大(或等于)，这个数就直接跳过；</span></span><br><span class="line"><span class="comment">//    反之如果当前数字比大根堆堆顶小，先poll掉堆顶，再将该数字放入堆中。</span></span><br><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span>&#123;</span><br><span class="line">    <span class="keyword">public</span> <span class="keyword">int</span>[] getLeastNumbers(<span class="keyword">int</span>[] arr, <span class="keyword">int</span> k) &#123;</span><br><span class="line">        <span class="keyword">if</span> (k == <span class="number">0</span> || arr.length == <span class="number">0</span>) &#123;</span><br><span class="line">            <span class="keyword">return</span> <span class="keyword">new</span> <span class="keyword">int</span>[<span class="number">0</span>];</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 默认是小根堆，实现大根堆需要重写一下比较器。</span></span><br><span class="line">        Queue&lt;Integer&gt; pq = <span class="keyword">new</span> PriorityQueue&lt;&gt;((v1, v2) -&gt; v2 - v1);</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> num: arr) &#123;</span><br><span class="line">            <span class="keyword">if</span> (pq.size() &lt; k) &#123;</span><br><span class="line">                pq.offer(num);</span><br><span class="line">            &#125; <span class="keyword">else</span> <span class="keyword">if</span> (num &lt; pq.peek()) &#123;</span><br><span class="line">                pq.poll();</span><br><span class="line">                pq.offer(num);</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        </span><br><span class="line">        <span class="comment">// 返回堆中的元素</span></span><br><span class="line">        <span class="keyword">int</span>[] res = <span class="keyword">new</span> <span class="keyword">int</span>[pq.size()];</span><br><span class="line">        <span class="keyword">int</span> idx = <span class="number">0</span>;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">int</span> num: pq) &#123;</span><br><span class="line">            res[idx++] = num;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> res;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>



<p><strong>3. 二叉搜索树也可以\</strong>O(NlogK)解决TopK问题～****    </p>
<p>​    BST相对没有前两种方法辣么热门，但是也很简单，和大根堆的思路差不多～不得不提的是，与前两种方法相比，<strong>BST的优势就是求得的前K个数字保证是有序的</strong>。</p>
<p>​    因为有重复的数字，所以用的是TreeMap而不是TreeSet（有的语言的标准库自带TreeMultiset，也是可以的）。TreeMap的key是数字，value是该数字的个数。我们遍历数组中的数字，维护一个数字总个数为K的TreeMap，每遍历一个元素：</p>
<p>​    1. 若目前map中数字个数小于K，则将map中当前数字对应的个数+1；</p>
<pre><code>2. 否则，判断当前数字与map中最大的数字的大小关系：若当前数字大于等于map中的最大数字，就直接跳过该数字；若当前数字小于map中的最大数字，则将map中当前数字对应的个数+1，并将map中最大数字对应的个数减1.</code></pre><figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span>&#123;</span><br><span class="line">    <span class="keyword">public</span> <span class="keyword">int</span>[] getLeastNumbers(<span class="keyword">int</span>[] arr, <span class="keyword">int</span> k) &#123;</span><br><span class="line">        <span class="keyword">if</span> (k == <span class="number">0</span> || arr.length == <span class="number">0</span>) &#123;</span><br><span class="line">            <span class="keyword">return</span> <span class="keyword">new</span> <span class="keyword">int</span>[<span class="number">0</span>];</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// TreeMap的key是数字, value是该数字的个数。</span></span><br><span class="line">        <span class="comment">// cnt表示当前map总共存了多少个数字。</span></span><br><span class="line">        TreeMap&lt;Integer, Integer&gt; map = <span class="keyword">new</span> TreeMap&lt;&gt;();</span><br><span class="line">        <span class="keyword">int</span> cnt = <span class="number">0</span>;</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> num: arr) &#123;</span><br><span class="line">            <span class="comment">// 1. 遍历数组，若当前map中的数字个数小于k，则map中当前数字对应个数+1</span></span><br><span class="line">            <span class="keyword">if</span> (cnt &lt; k) &#123;</span><br><span class="line">                map.put(num, map.getOrDefault(num, <span class="number">0</span>) + <span class="number">1</span>);</span><br><span class="line">                cnt++;</span><br><span class="line">                <span class="keyword">continue</span>;</span><br><span class="line">            &#125; </span><br><span class="line">            <span class="comment">// 2. 否则，取出map中最大的Key（即最大的数字), 判断当前数字与map中最大数字的大小关系：</span></span><br><span class="line">            <span class="comment">//    若当前数字比map中最大的数字还大(或等于)，就直接忽略；</span></span><br><span class="line">            <span class="comment">//    若当前数字比map中最大的数字小，则将当前数字加入map中，并将map中的最大数字的个数-1。</span></span><br><span class="line">            Map.Entry&lt;Integer, Integer&gt; entry = map.lastEntry();</span><br><span class="line">            <span class="keyword">if</span> (entry.getKey() &gt; num) &#123;</span><br><span class="line">                map.put(num, map.getOrDefault(num, <span class="number">0</span>) + <span class="number">1</span>);</span><br><span class="line">                <span class="keyword">if</span> (entry.getValue() == <span class="number">1</span>) &#123;</span><br><span class="line">                    map.pollLastEntry();</span><br><span class="line">                &#125; <span class="keyword">else</span> &#123;</span><br><span class="line">                    map.put(entry.getKey(), entry.getValue() - <span class="number">1</span>);</span><br><span class="line">                &#125;</span><br><span class="line">            &#125;</span><br><span class="line">            </span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 最后返回map中的元素</span></span><br><span class="line">        <span class="keyword">int</span>[] res = <span class="keyword">new</span> <span class="keyword">int</span>[k];</span><br><span class="line">        <span class="keyword">int</span> idx = <span class="number">0</span>;</span><br><span class="line">        <span class="keyword">for</span> (Map.Entry&lt;Integer, Integer&gt; entry: map.entrySet()) &#123;</span><br><span class="line">            <span class="keyword">int</span> freq = entry.getValue();</span><br><span class="line">            <span class="keyword">while</span> (freq-- &gt; <span class="number">0</span>) &#123;</span><br><span class="line">                res[idx++] = entry.getKey();</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> res;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

<h2 id="Python"><a href="#Python" class="headerlink" title="Python"></a>Python</h2><p><strong>Solution :</strong></p>
<figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line"></span><br><span class="line"></span><br></pre></td></tr></table></figure>




      
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